Q 4.39 - Intermediate Algebra

Question

Translate to a system of equations and then solve:

Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs $155$ feet of fencing to enclose the pool. The length of the long side is $10$ feet less than twice the width. Find the length and width of the pool area to be enclosed.

Step-by-Step Solution

Answer

The length and width of the pool area is $60ft$ and $35ft$.

Step 1 . Given

We are given that the length of the long side is $10$ feet less than twice the width and Mario needs $155ft$ fencing.

Step 2 . Assumptions and formation of equations.

Let $l$ be the length of long side and $w$ be the length of small side,

So, the perimeter of the pool is given by $2l+w=155$.

And the length of the long side is $10$ feet less than twice the width, so

$l=2w-10$

Step 3 . Solving the equation.

Putting the value of $l$ in first equation, we get

$$\begin{gathered} 2(2w-10)+w=155 \\ 4w-20+w=155 \\ 5w=155+20 \\ 5w=175 \end{gathered}$$

Dividing by $5$, we get

$$\begin{gathered} w=\frac{175}{5} \\ w=35ft \end{gathered}$$

Now, putting the value $w$ in second equation,

$$\begin{gathered} l=2\times 35-10 \\ l=70-10 \\ l=60ft \end{gathered}$$

Step 4. Checking the solution

Checking the solution by putting the value of $l,w$ in the equations, we get

$$\begin{gathered} 2l+w=155 \\ 2\left(60\right)+35=155 \\ 120+35=155 \\ 155=155 \\ l=2w-10 \\ 60=2\left(35\right)-10 \\ 60=70-10 \\ 60=60 \end{gathered}$$

This is true, hence the solution is correct.